答案:http://www.pudn.com/downloads36/sourcecode/app/detail115926.html专门针对你说的写了一个简单点的程序你看看可以用不???// 学生成绩管理系统.cpp : Defines the entry point for the console application.///*求C++课程设计答案,望高手相助 悬赏分:50 - 离问题结束还有 17 天 5 小时 设计一个学生类,包括数据成员姓名、分数、等级(字符型数组),设计友元函数,用来根据成绩分数确定对应的等级:90~100:优; 80~89:良; 70~79:及格; 60以下:不及格; disp()成员函数用来显示学生的数据。主函数中定义学生类对象数组,初始化姓名和分数;输出结果;显示每个学生的姓名、分数、等
其他:最近都是课程实践的问题 哎~~~
看看能不能用
#include <stdio.h>
#include <string.h>
const int MAX_NUM = 5;
const int MAX_NAME_LEN = 127;
const int MAX_SUBJECT = 3;
const double SUBJECTS = 3.0;
typedef struct _Student
{
char szName[MAX_NAME_LEN + 1];
int nArryScore[MAX_SUBJECT];
int nSum;
double dAverage;
}Student;
void Sort(Student students[], int n)
{
Student stTmp;
int nTmp = 0;
int nPos = 0;
int nIndex = 0;
for (nPos = 0; nPos < n - 1; nPos++)
{
nTmp = nPos;
memset(&stTmp, 0, sizeof(Student));
for (nIndex = nPos + 1; nIndex < n; nIndex++)
{
if (students[nIndex].nSum > students[nTmp].nSum)
{
nTmp = nIndex;
}
}
memcpy(&stTmp, &students[nTmp], sizeof(Student));
memcpy(&students[nTmp], &students[nPos], sizeof(Student));
memcpy(&students[nPos], &stTmp, sizeof(Student));
}
}
int main (void)
{
printf("请输入五个同学三门课的成绩:\n");
Student students[MAX_NUM];
memset(students, 0, sizeof(students));
Student* pStudent = NULL;
int nCnt = 0;
int nPos = 0;
while (MAX_NUM > nCnt)
{
printf("第%d个同学的记录:", nCnt + 1);
pStudent = &students[nCnt];
scanf("%s %d %d %d", pStudent->szName,
&pStudent->nArryScore[0],
&pStudent->nArryScore[1],
&pStudent->nArryScore[2]);
for (nPos = 0; nPos < MAX_SUBJECT; nPos++)
{
pStudent->nSum += pStudent->nArryScore[nPos];
}
pStudent->dAverage = pStudent->nSum / SUBJECTS;
nCnt++;
getchar();
}
Sort(students, MAX_NUM);
printf("------------------------------------------------\n");
printf("输出排序后五个同学的三门成绩:\n");
printf("------------------------------------------------\n");
printf("排序 姓名 课1 课2 课3 总分 平均分\n");
for (nPos = 0; nPos < MAX_NUM; nPos++)
{
pStudent = &students[nPos];
printf("第%d名: %s %d %d %d %d %.2f\n", nPos + 1,
pStudent->szName,
pStudent->nArryScore[0],
pStudent->nArryScore[1],
pStudent->nArryScore[2],
pStudent->nSum,
pStudent->dAverage);
}
return 0;
}
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